Science:Math Exam Resources/Courses/MATH102/December 2014/Question C 02 (d)/Solution 1

Same as in C2(a), to solve the steady states, we just need to solve the following equation $c(rc^{2}-P_{m}c+k^{2}r)=0,$ Obviously $c=0$ is one of the steady states and other steady states are roots of the quadratic equation $rc^{2}-P_{m}c+k^{2}r=0.$ To get only two steady state, from the quadratic formula we need to have $b^{2}-4ac=0$ and in this case $(P_{m})^{2}-4\cdot r\cdot k^{2}r=0.$ Thus with the given values of $P_{m}$ and $k$ we have $r^{2}={\frac {P_{m}^{2}}{4k^{2}}}={\frac {1}{16}},$ To obtain positive steady state $c$, we get $r={\frac {1}{4}}.$